Electrochemistry
CollegeElectron transfer as a source of work
Electrochemistry studies reactions involving electron transfer — oxidation (loss of electrons, increase in oxidation number) and reduction (gain of electrons, decrease in oxidation number) — and harnesses them to do electrical work. A galvanic (voltaic) cell separates oxidation and reduction into two half-cells connected by a wire (external electron path) and a salt bridge (ion path to maintain charge balance), converting a spontaneous redox reaction into usable current. An electrolytic cell does the reverse: it uses an external power source to force a non-spontaneous redox reaction to occur (electrolysis).
Each half-reaction has a standard reduction potential, E°, measured against the reference hydrogen electrode (defined as 0 V). The overall cell potential is:
A positive E°cell indicates a spontaneous reaction (ΔG° < 0), linked by ΔG° = −nFE°cell (F = Faraday's constant, 96 485 C mol⁻¹; n = moles of electrons transferred). Under non-standard conditions, the Nernst equation corrects the potential:
Why a battery goes flat, and how electroplating works
A battery is a galvanic cell: it produces current only as long as the spontaneous redox reaction has somewhere to go. As the cell operates, reactant concentrations at each electrode shift — the Nernst equation shows this directly, since as Q moves away from 1 (products building up, reactants depleting), Ecell drops. When the system reaches equilibrium, Q = K, Ecell = 0, and the "flat" battery can no longer do work — even though reactants may not be completely gone, the driving force has vanished.
Electrolysis runs this logic backward: because plating a metal onto an object (e.g. chrome-plating a bumper) is often non-spontaneous on its own, an external power supply forces electrons where they wouldn't naturally go, driving reduction of metal ions onto the object being plated as the cathode. The amount of metal deposited is quantitatively predictable from the current and time via Q = It and Faraday's laws — connecting electrical charge directly to moles of electrons and hence moles of metal deposited.
"Red Cat, An Ox" — REDuction occurs at the CAThode; oxidation occurs at the ANode. True for both galvanic and electrolytic cells, even though the sign of the electrode's charge flips between the two cell types.
Worked example — cell potential and spontaneity
A cell is built from Zn²⁺/Zn (E° = −0.76 V) and Cu²⁺/Cu (E° = +0.34 V) half-cells. Determine E°cell, identify the cathode and anode, and state whether the reaction is spontaneous.
- Compare the two standard reduction potentials: Cu²⁺/Cu (+0.34 V) is higher than Zn²⁺/Zn (−0.76 V), so Cu²⁺ is reduced (it is the stronger oxidising agent) and Zn is oxidised.
- Assign electrodes: reduction happens at the cathode (Cu), oxidation at the anode (Zn).
- Apply the formula: E°cell = E°cathode − E°anode = 0.34 − (−0.76).
- Compute: E°cell = +1.10 V.
- Since E°cell > 0, the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is spontaneous as written — this is the classic Daniell cell.
- Confirm with ΔG°: n = 2 electrons transferred, ΔG° = −nFE°cell = −2 × 96485 × 1.10 = −212 267 J mol⁻¹ ≈ −212 kJ mol⁻¹, strongly negative, consistent with spontaneity.
In the Daniell cell above, which electrode loses mass over time, and why?
The zinc anode loses mass. Zinc is oxidised, Zn(s) → Zn²⁺(aq) + 2e⁻, so solid zinc atoms leave the electrode and enter solution as ions, while at the copper cathode, Cu²⁺ ions are reduced and deposit as solid copper, so the copper electrode gains mass.
Use the Nernst equation to find Ecell for the Daniell cell at 25 °C when [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M (E°cell = 1.10 V, n = 2).
Q = [Zn²⁺]/[Cu²⁺] = 1.0/0.010 = 100. E = E° − (0.0592/n)log Q = 1.10 − (0.0592/2)log(100) = 1.10 − (0.0296 × 2) = 1.10 − 0.0592 = 1.04 V. Lower Cu²⁺ concentration reduces the driving force slightly, as expected.
Why does electrolysis require an external power source while a galvanic cell does not?
A galvanic cell houses a spontaneous redox reaction (ΔG° < 0, E°cell > 0) that proceeds on its own, releasing energy as electrical work. Electrolysis drives a non-spontaneous reaction (ΔG° > 0, E°cell < 0 as written) "uphill," which requires an external EMF exceeding the cell's own opposing potential to force electrons in the non-favoured direction.
How many grams of copper are deposited by a current of 2.00 A flowing for 1 hour through a Cu²⁺ solution? (M(Cu) = 63.5 g mol⁻¹, F = 96 485 C mol⁻¹, Cu²⁺ + 2e⁻ → Cu.)
Charge Q = It = 2.00 × 3600 = 7200 C. Moles of electrons = Q/F = 7200/96485 = 0.0746 mol e⁻. Since 2 mol e⁻ deposit 1 mol Cu, moles Cu = 0.0746/2 = 0.0373 mol. Mass = 0.0373 × 63.5 = 2.37 g of copper.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1A cell has E cathode = +0.34 V (Cu) and E anode = -0.76 V (Zn). Find E cell.
- E cell = E cathode - E anode.
- E cell = 0.34 - (-0.76).
- Compute: 0.34 + 0.76 = 1.10 V.
Example 2Is a cell with E cell = -0.45 V spontaneous?
- A cell is spontaneous only if E cell is positive.
- Here E cell = -0.45 V, which is negative.
- So the forward reaction does not run on its own.
Now you try
Work each one out first, then tap to reveal the worked answer.